Question

4tanA%1-tan^4A=tan2A+sin2A​

Answer 1

Answer:

See prove of the identity below

Step-by-step explanation:

Recall that the following difference can be factored out as:

$(1-tan^4(A)) = (1+tan^2(A)) *(1-tan^2(A))$

Now we look at the right side of the equal sign, and recall the identity that relates the sine of a double angle 2A with an expression in terms of the tan(A):

$sin(2A) = \frac{2\,tan(A)}{1+tan^2(A)}$

and similarly, the tangent of the double angle "2A" can be written as:

$tan (2A) = \frac{2*tan(A)}{1-tan^2(A)}$

We can then combine the two expressions on the right using  the common denominator: $(1-tan^4(A)) = (1+tan^2(A)) *(1-tan^2(A))$:

$tan(2A) +sin(2A)=\frac{2*tan(A)*(1+tan^2(A))}{1-tan^4(A)} +\frac{2*tan(A)*(1-tan^2(A))}{1-tan^4(A)} \\=\frac{2*tan(A)+2*tan(A)}{1-tan^4(A)} =\frac{4*tan(A)}{1-tan^4(A)}$

And we have proved the identity since the right hand side becomes exactly equal to the left hand side of the equal sign.